Max Dot Product of Two Subsequences | Leetcode 1458

Time Complexity : O(n*m)
Space Complexity : O(n*m)
Problem Link : leetcode.com/p...
Python Code Link: github.com/Ayu...
Please like, share and subscribe if you found the video useful. Feel free to ask in comments section if you have any doubts. :)
#DataStructuresAndAlgorithms
#Leetcode
#interviewpreparation
#AyushiSharma
Max Dot Product of Two Subsequences solution
Max Dot Product of Two Subsequences Leetcode
Max Dot Product of Two Subsequences C++
Max Dot Product of Two Subsequences Java
Max Dot Product of Two Subsequences Python
🔥🔥🔥🔥👇👇👇
Join telegram channel for more updates on placement preparation : t.me/AyushiSha...
Checkout the series: 🔥🔥🔥
👉Interview Experiences : • Interview Experiences
👉 Array: • Arrays
👉 Linked List : • Linked List
👉 Heap : • Heap
👉 Recursion : • Recursion
👉 Stack and Queue : • Stack And Queue
👉 Greedy : • Greedy
👉 Dynamic Programming : • Dynamic Programming
👉 Leetcode contests : • Leetcode contests
👉 Leetcode June Challenge : • Leetcode June Challenge
👉 Leetcode July Challenge : • Leetcode July Challenge
LIKE | SHARE | SUBSCRIBE 🔥🔥😊

[@vaibhavgupta973]

thx

[@AyushiSharmaDSA]

Welcome 😊

[@bharatrajpurohit5300]

Nice Explanation

[@AyushiSharmaDSA]

thank you :)

[@vision7673]

great explanation ma'am

[@AyushiSharmaDSA]

thank you :)

[@pritz9]

Nice Explanation !

[@AyushiSharmaDSA]

Thanks Prithwish😊

[@subendernath]

I've been recommending your videos to all my coding buddies. Keep 'em coming! 😊

[@AyushiSharmaDSA]

Thank you Subender 😊

[@ApnaVlogs-tj7do]

Optimised Solution ( Tabulation Approach ) 👇 class Solution: def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int: n = len(nums1) m = len(nums2) mini_num1 = 1e9 maxi_num1 = -1e9 mini_num2 = 1e9 maxi_num2 = -1e9 for num in nums1: if num > maxi_num1: maxi_num1 = num if num < mini_num1: mini_num1 = num for num in nums2: if num > maxi_num2: maxi_num2 = num if num < mini_num2: mini_num2 = num if (maxi_num1 < 0 and mini_num2 > 0) or (maxi_num2 < 0 and mini_num1 > 0): return max(maxi_num1*mini_num2,mini_num1*maxi_num2) dp = [[0]*(m+1) for i in range(n+1)] for ind1 in range(1,n+1): for ind2 in range(1,m+1): first = nums1[ind1-1] * nums2[ind2-1] + dp[ind1-1][ind2-1] second = 1 * dp[ind1][ind2-1] third = 1 * dp[ind1-1][ind2] dp[ind1][ind2] = max(first,second,third) return dp[n][m]

[@Abhishek-ys2io]

Finally video after long time.. why?

[@vamsimadugula8524]

we can also consider the case like we can not take both i and j also and making sure at last non empty ??

[@ApnaVlogs-tj7do]

challenging part is to figure out how to handle the base case for [-1,-1] & [1,1] . rest is easy

[@brawlstarsac3574]

please explain in English

[@callme_SRK]

Mam can I get ur insta ID or mail ID?

[@AyushiSharmaDSA]

Sure [e-mail]

[@tofpgaming666]

and password @@AyushiSharmaDSA